Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises: 40


$f'(x) = \dfrac{-9\cot ^8 \theta \csc ^2 \theta}{2\sqrt{\cot ^9 \theta +1}}$

Work Step by Step

In order to derivate this function you have to apply the chain rule Let's make an «u» substitution to make it easier $u = \cot ^9 \theta +1$ $f(u) = \sqrt{u} $ Derivate the function: $f'(u) = \dfrac{u'}{2\sqrt{u}} $ Now let's find u' $u' = -9\cot ^8 \theta \csc ^2 \theta$ Then undo the substitution, simplify and get the answer: $f'(x) = \dfrac{-9\cot ^8 \theta \csc ^2 \theta}{2\sqrt{\cot ^9 \theta +1}}$
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